Question

(2) Assume A = 1 1 -1]: [10 11 (a) Find the maximum and minimum of ||A7|| subject to ||2|| = 2. (b) Classify the quadratic form Q1(x) = xAT Ax. (c) Classify the quadratic form Q2(x) = xT AATX. (d) Find a singular value decomposition for A.

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Answer 1

Answer: The materia Given that ACT! 1412 - 122003 lla! 9 2 2 2 +1 +1 +1 = (1+1+1+1+1% =153' || All= VS (a) from the given an ca) 1311 = 22 1. The maximum of 11 All E||A11 112 11 efs x2 sars - max. of 11Aall = 25 The minimum of 1ARI1 = 11A11 11211 25 x2 min. of 11Aill za

(6) The qua doratie form of 6, (x) = A Ax , A A = 1 li l -1 Aso (1 -1 trillio7 11-111!!! Til 0+1 1-17 ofl ofl 0-1 | TYL. 0-1 1+1 na 1 07 = x AA 2,(8) = x AAX x = (xxxits], x=( ( 307 O A AAX 10-12) 1 X3 > [au, the M44-13-X,+22] (2 - axét x, X2 + x, X2+x2-x2- X5X3+2x+ tax = 2*,* + x2 +0%+ +8X,X2 - 2X, X3

© The quebratic form 0,60) = xTAATX AAT. [10171117 JIJ ritori 1 -17 (140-1 HIHI 11 {axt swit] 20 .6) = xa'- ex?+372 AA

& Given that The mandarin A=( 10 ) To finding singular value decomposition super finest to finding Ata 0 Nthala101 step-2; coola To find eigen values are (A-AI/=0 LATA - 1] /=0 1 0-2 10 , i ind-/ leo Lo - 2-d1 (2-14(1-(2-)-1)-1(2-1) =0 (0-1) (0-1) (2-] -2-1) -(2-1)=0 (9-x)(1-) – 21a-)0 (8-) ((2-1) (1-)-2)=0 (9-x) = 0, $-ax-x+x² -&zo

12-31-0, 2-4=0 1(4-3)=0,8-1=0 130, 133, 1:2. Therefore the Eigen values are, Step-3,30, 12 = 2, 13=3. Then finding eigen vetens fist to find Ants (1-3 ) - (28]=usut IMAT 101 - 12-1,0/-0 (2-2) (3-4) -0 =0 (2-1)(3-1) = 0 2.52, 4:3 Eigenvector corresponding to e aoill !I - 102,T - AAT = 102 103 ro07 Je,+C2 > n Value 1201-1207

- The Eigen vector coresponding value is li voilippe The eigen rector corresponding Eigenvalue's is 11-6053-1951-663) 16] -(0] cocitla - The Eigen vector coresponding segenvalue is